We say a (singular) integral curve in \(\PP^n\) is strange if there is a point which lies on all the tangent lines at nonsingular points of the curve.
There are many singular strange curves, e.g., the curve given parametrically by \(x=t, y=t^p, z=t^{2p}\) over a field of characteristic \(p>0\).
Show, however, that if \(\characteristic k=0\), there aren’t even any singular strange curves besides \(\PP^1\).
Solution
1. In characteristic \(p>0\), the projective closure of the parametrizedcurve is the image of\[ \nu:\PP^1\longrightarrow\PP^3, \]\[ [u:v] \longmapsto \bigl[ u v^{2p-1}: u^p v^p: u^{2p}: v^{2p} \bigr]. \]
The four displayed forms are homogeneous of degree \(2p\) and have no common zero, so they define a morphism on all of \(\PP^1\). On the chart \(v\ne0\), putting \[ t=\frac uv \] and dividing by \(v^{2p}\) gives \[ [t:t^p:t^{2p}:1]. \] Thus the image contains the affine parametrized curve \[ x=t,\qquad y=t^p,\qquad z=t^{2p}, \] and, being projective, is its projective closure. The first affine coordinate recovers \(t\), so the map is birational onto its integral image.
2. The affine part of this curve is nonsingular, and its unique point atinfinity is\[ Q=[0:0:1:0]. \]
On the affine chart \(w=1\), the image is the graph \[ y=x^p, \qquad z=x^{2p}, \] whose coordinate ring is \[ k[x,y,z]/(y-x^p,z-x^{2p}) \cong k[x]. \] Hence the affine part is isomorphic to \(\AA^1\) and is nonsingular.
1. At the unique point\([u:v]=[1:0]\), the image is\[ Q=[0:0:1:0]. \]Thus \(Q\) is the only point at infinity.
3. The point \(Q\) is singular.
Work on the chart \(z\ne0\) around \(Q\). Put \[ s=\frac vu. \] Dividing the parametrization in step 1 by \(u^{2p}\) gives \[ \frac{x}{z}=s^{2p-1}, \qquad \frac{y}{z}=s^p, \qquad \frac{w}{z}=s^{2p}. \] Thus the local ring of the image at \(Q\) is the localization at its origin of \[ A=k[s^p,s^{2p-1},s^{2p}] \subseteq k[s]. \] Let \(\mathfrak m\) be its maximal ideal. The classes of \[ s^p \qquad\text{and}\qquad s^{2p-1} \] are linearly independent in \[ \mathfrak m/\mathfrak m^2: \] every nonzero element of \(\mathfrak m^2\) has \(s\)-adic order at least \(2p\), whereas these two elements have orders \(p\) and \(2p-1\). Therefore \[ \dim_k\mathfrak m/\mathfrak m^2\ge2. \] The local ring has dimension \(1\), so it is not regular. Hence \(Q\) is singular.
4. Every tangent line at a nonsingular point of the curve passes throughthe fixed point\[ A=[1:0:0:0]. \]
By step 2, every nonsingular point lies on the affine chart \(w=1\) and has the form \[ P_t=(t,t^p,t^{2p}). \] Differentiating the affine parametrization gives \[ \frac d{dt}(t,t^p,t^{2p}) = (1,pt^{p-1},2pt^{2p-1}) = (1,0,0), \] because \(\operatorname{char}k=p\). Thus the affine tangent line at \(P_t\) is \[ \lambda \longmapsto (t+\lambda,t^p,t^{2p}). \] Its projective closure contains the point at infinity in the \(x\)-direction, namely \[ A=[1:0:0:0], \] independently of \(t\). Therefore the curve is strange. Together with step 3, this proves (a).
5. Now assume \(\operatorname{char}k=0\). Let\[ C\subseteq\PP^n \]be an integral strange curve, and let\[ A\in\PP^n \]lie on every tangent line at a nonsingular point of \(C\). Projection from\(A\) has zero differential at every nonsingular point where it is defined.
Choose projective coordinates with \[ A=[1:0:\cdots:0]. \] Projection from \(A\) is \[ \pi_A: [x_0:x_1:\cdots:x_n] \dashrightarrow [x_1:\cdots:x_n]. \] It is defined on \[ C\setminus\{A\}. \]
Let \(P\) be a nonsingular point in this open set. The differential of \(\pi_A\) kills exactly the tangent direction of the line through \(A\) and \(P\): infinitesimally, moving along that line changes only the coordinate discarded by the projection. By strangeness, the tangent line \(T_PC\) is precisely a line through \(A\). Since \(T_PC\) is one-dimensional, \[ d\pi_A|_{T_PC}=0. \]
6. If the projection of \(C\) from \(A\) were nonconstant, step 5 wouldcontradict characteristic zero.
Suppose the rational projection is nonconstant, and let \[ Y \subseteq \PP^{n-1} \] be its integral image. Shrink to dense nonsingular opens \[ U\subseteq C\setminus\{A\}, \qquad V\subseteq Y, \] such that \[ \pi:U\longrightarrow V \] is a morphism of smooth curves. The induced extension of function fields \[ k(Y)\subseteq k(C) \] is finite. Because \(\operatorname{char}k=0\), it is separable.
For a finite separable extension of one-variable function fields, the map on absolute differentials \[ k(C)\tensor_{k(Y)}\Omega_{k(Y)/k} \longrightarrow \Omega_{k(C)/k} \] is an isomorphism. Indeed, the transitivity sequence has \[ \Omega_{k(C)/k(Y)}=0 \] for a finite separable extension, while both remaining vector spaces have dimension \(1\) over \(k(C)\).
Hence the differential of \(\pi\) at the generic point is nonzero. On the other hand, step 5 says that the differential vanishes at every nonsingular closed point of \(U\). The induced morphism of line bundles \[ \pi^*\Omega_{V/k} \longrightarrow \Omega_{U/k} \] therefore vanishes identically, and in particular vanishes at the generic point. This is a contradiction. Thus projection from \(A\) cannot have one-dimensional image.
7. The curve \(C\) is a line.
By step 6, projection from \(A\) is constant on the dense open where it is defined. A fibre of linear projection from \(A\) is a line through \(A\). Therefore a dense open subset of \(C\) lies on one fixed line \[ L\subseteq\PP^n. \] Taking closures gives \[ C\subseteq L. \] Both are integral projective curves, so \[ C=L\cong\PP^1. \] Thus in characteristic zero the only strange integral curve is a line, and in particular there are no singular strange curves. This proves (b).
8. Q.E.D.
Steps <1>1–<1>4 prove the positive-characteristic example in (a). Steps <1>5–<1>7 prove that in characteristic zero every strange integral curve is a line, proving (b).